Question

$\sf \to \: \sf \: find \: \dfrac{dy}{dx} \: \: \: if \: \: y = (tanx) ^{sinx} + (sinx)^{tanx} \: \: w.r.t \: \: x$
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Answer 1

Answer:-

let $\rm\\\\\\ \rm (tanx) ^{sinx} \sf{ \: be \: } y_1$

and $\rm\\\\\\\rm (sinx)^{tanx} \sf{\: be \:} y_2$

$\rm$

$\rm \bullet \bf (tanx) ^{sinx} = y_1$

taking log on both sides

$\rm \log \ y_1 = \log \: ( tanx)^{sinx}$

$\rm \log \ y_1 = sinx \cdot \ln \: tanx$

$\rm \dfrac{1}{y_1} y_1 ' = \sin x \dfrac{1}{\tan x} sec^2 x+ \log \tan x \cdot \cos x$

$\rm$

$\rm \bullet \bf(sinx)^{tanx} = y_2$

taking log on both sides

$\rm \log \ y_2 = \log \: ( sinx)^{tanx}$

$\rm \log \ y_2 = tanx \cdot log \ sinx$

$\rm \dfrac{1}{y_2} y_2 ' = tanx \cdot \dfrac{1}{sinx} \cdot cosx + \log sinx \ . \ sec^2 x$

$\rm$

now to find dy/dx, we need to add 1/y₁ y₁' and 1/y₂ y₂'

$\rm = \sin x \dfrac{1}{\tan x} sec^2 x+ \log \tan x \cdot \cos x$

$\rm + tanx \cdot \dfrac{1}{sinx} \cdot cosx + \log sinx \ . \ sec^2 x$

$\rm = secx + sec^2 x \ log(sinx) + cosx \: log(tanx) +1$

Answer 2

EXPLANATION.

$\sf \implies \: y \: = ( \tan(x) ) {}^{ \sin(x) } \: + \: (\sin(x)) {}^{ \tan(x) }$

$\sf \implies \:let \: y \: = u \: + v \\ \\ \sf \implies \: \frac{dy}{dx} = \frac{du}{dx} \: + \: \frac{dv}{dx} \\ \\ \sf \implies \: \: u \: = (\tan(x)) {}^{ \sin(x) } \\ \\ \sf \implies \: \: v \: =( \sin(x)) {}^{ \tan(x) }$

$\sf \implies \:u \: = (\tan(x)) {}^{ \sin(x) } \\ \\ \sf \implies \: \: taking \: log \: on \: both \: sides \: we \: get \\ \\ \sf \implies \: ln(u) = ln( \tan(x) ) {}^{ \sin(x) } \\ \\ \sf \implies \: \frac{1}{u} \frac{du}{dx} = \sin(x) ln( \tan(x) )$

$\sf \implies \: \dfrac{1}{u} \dfrac{du}{dx} = \sin(x) . \dfrac{ \sec {}^{2} (x) }{ \tan(x) } \: + \: ln( \tan(x) ) \cos(x) \\ \\ \sf \implies \: \frac{du}{dx} = u \Bigg[ \sec(x) \: + \: ln( \tan(x) ) \cos(x) \Bigg] \\ \\ \sf \implies \: \frac{du}{dx} \: = (\tan(x) ) {}^{ \sin(x) }\Bigg[ \sec(x) \: + ln( \tan(x) ) \cos(x) \Bigg]$

$\sf \implies \: v \: =( \sin(x) ) {}^{ \tan(x) } \\ \\ \sf \implies \: taking \: log \: on \: both \: sides \: we \: get \\ \\ \sf \implies \: \frac{1}{v}. \frac{dv}{dx} = ln( \sin(x) ) {}^{ \tan(x) } \\ \\ \sf \implies \: \frac{1}{v} . \frac{dv}{dx} = \tan(x) ln( \sin(x) )$

$\sf \implies \: \dfrac{1}{v}. \dfrac{dv}{dx} = \tan(x). \dfrac{ \cos(x) }{ \sin(x) } \: + \: ln( \sin(x) ) \sec {}^{2} (x) \\ \\ \sf \implies \: \frac{dv}{dx} = v \Bigg[1 \: + \: ln( \sin(x) ) \sec {}^{2} (x) \Bigg] \\ \\ \sf \implies \: \frac{dv}{dx} \: = \sin(x) {}^{ \tan(x) }\Bigg[1 \: + \: ln( \sin(x) ) \sec {}^{2} (x) \Bigg]$

$\sf \implies \: y \: = u \: + v \: \\ \\ \sf \implies \: add \: the \: value \: of \: u \: and \: v \\ \\ \sf \implies \: \tan(x) {}^{ \sin(x) }\Bigg[ \sec(x) \: + \: ln( \tan(x) ) \cos(x) \Bigg] \: + \: \sin(x) {}^{ \tan(x) } \Bigg[ 1 \: + \: ln( \sin(x) ) \sec {}^{2} (x) \Bigg]$